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I have the following problem:

Let $M$ be a connected closed $7$-manifold such that $H_1(M,\mathbb{Z}) = 0$, $H_2(M,\mathbb{Z}) = \mathbb{Z}$, $H_3(M,\mathbb{Z}) = \mathbb{Z}/2$. Compute $H_i(M,\mathbb{Z})$ and $H^i(M,\mathbb{Z})$ for all $i$.

I know that if $M$ is orientable, using Poincaré duality, the fact that $\chi(M)=0$ and the exact sequence for $H^i(M,\mathbb{Z})$ I can get the result.

But, I don't know how to prove that $M$ is orientable. I know that is the case if $M$ does not have $2$-torsion on $\pi_1(M)$, but I don't see why this $2$-torsion should descend to $H_1(M)$.

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  • $\begingroup$ Yeah, sorry I will edit. $\endgroup$ – abrax Jul 19 '16 at 5:31
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    $\begingroup$ Just to point out the connection between the two answers: $Hom(\pi_1(M),\mathbb Z/2)\cong Hom(H_1(M),\mathbb Z/2)\cong H^1(M,\mathbb Z/2)$. Eric talks about a distinguished element on the left side, which under the natural isomorphisms corresponds to $\omega_1(M)$ on the right side, which is the side which Najib uses for his argument. $\endgroup$ – Daniel Valenzuela Jul 19 '16 at 14:19
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For any connected manifold $M$, there is a homomorphism $\pi_1(M)\to\mathbb{Z}/2$ which sends a loop to $0$ if going around the loop preserves orientation and sends the loop to $1$ if going around the loop reverses orientation. This homomorphism is trivial iff $M$ is orientable. Since $\mathbb{Z}/2$ is abelian, this homomorphism factors through the Hurewicz map $\pi_1(M)\to H_1(M)$. In particular, this means that if $H_1(M)=0$, the homomorphism is trivial so $M$ is orientable.

(By the way, the statement that $\chi(M)=0$ does not require $M$ to be orientable--you can prove it using mod $2$ Poincare duality, for instance.)

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A quick proof using Stiefel–Whitney classes: a manifold $M$ is orientable iff the first SW class $w_1(M) \in H^1(M;\mathbb{Z}/2\mathbb{Z})$ is zero. But by the universal coefficient theorem, $$H^1(M;\mathbb{Z}/2\mathbb{Z}) = \operatorname{Hom}_\mathbb{Z}(H_1(M;\mathbb{Z}), \mathbb{Z}/2\mathbb{Z}) = 0.$$

Of course under the hood I don't think there's anything more than what you can find in Eric Wofsey's answer, but this argument is quite simple and shows the power of characteristic classes.

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